Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 4}{x + 8} = \dfrac{-5x + 28}{x + 8}$
Multiply both sides by $x + 8$ $ \dfrac{x^2 + 4}{x + 8} (x + 8) = \dfrac{-5x + 28}{x + 8} (x + 8)$ $ x^2 + 4 = -5x + 28$ Subtract $-5x + 28$ from both sides: $ x^2 + 4 - (-5x + 28) = -5x + 28 - (-5x + 28)$ $ x^2 + 4 + 5x - 28 = 0$ $ x^2 - 24 + 5x = 0$ Factor the expression: $ (x - 3)(x + 8) = 0$ Therefore $x = 3$ or $x = -8$ However, the original expression is undefined when $x = -8$. Therefore, the only solution is $x = 3$.